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klasskru [66]
3 years ago
15

Seismographs measure the arrival times of earthquakes with a precision of 0.125 s. To get the distance to the epicenter of the q

uake, they compare the arrival times of S- and P-waves, which travel at different speeds. If S- and P-waves travel at 3.74 and 7.22 km/s, respectively, in the region considered, how precisely can the distance to the source of the earthquake be determined
Physics
1 answer:
zloy xaker [14]3 years ago
4 0

Answer:

435 m

Explanation:

The precision with which the distance to the source of the earthquake can be estimated is equal to the difference in distance covered by the S- and P-waves in the time of 0.125 s.

The distance covered by each type of wave is given by

d=vt

where

v is the speed of the waves

t is the time

For S-waves,

v = 3.74 km/s

t = 0.125 s

So the distance covered is

d_s=(3.74)(0.125)=0.4675 km = 467.5 m

For P-waves,

v = 7.22 km/s

t = 0.125 s

So the distance covered is

d_p=(7.22)(0.125)=0.9025 km = 902.5 m

So, the precision with which the distance can be determined is:

\Delta d = 902.5 - 467.5 =435 m

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In this case, the absence of external forces will make the pucks move in the form of a uniform circular motion.

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The landing of a spacecraft is cushioned with the help of airbags. During its landing on Mars, the velocity of downward fall is
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3 years ago
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A helicopter starting on the ground is rising directly into the air at a rate of 25 ft/s. You are running on the ground starting
rusak2 [61]

Answer:

The rate of change of the distance between the helicopter and yourself (in ft/s) after 5 s is \sqrt{725} ft/ sec

Explanation:

Given:

h(t) =  25 ft/sec

x(t) = 10 ft/ sec

h(5) = 25 ft/sec . 5 = 125 ft

x(5) = 10 ft/sec . 5 = 50 ft

Now we can calculate the distance between the person and the helicopter by using the Pythagorean theorem

D(t) = \sqrt{h^2 + x^2}

Lets find the derivative of distance with respect to time

\frac{dD}{dt} (t)  = \frac{2h \cdot \frac{dh}{dt} +2x \cdot\frac{dx}{dt}} {2\sqrt{h^2 + x^2}}

Substituting the values of h(t) and  x(t) and simplifying we get,

\frac{dD}{dt}(t) = \frac{50t \cdot \frac{dh}{dt} + 20 \cdot \frac{dx}dt}{2\sqrt{625\cdot t^2 + 100 \cdot t^2}}

\frac{dh}{dt} = 25ft/sec

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5 0
3 years ago
(a) Calculate the absolute pressure at the bottom of a freshwater lake at a point whose depth is 30.0 m. Assume the density of t
Law Incorporation [45]

Answer:

(a) The absolute pressure at the bottom of the freshwater lake is 395.3 kPa

(b) The force exerted by the water on the window is 36101.5 N

Explanation:

(a)

The absolute pressure is given by the formula

P = P_{o} + \rho gh

Where P is the absolute pressure

P_{o} is the atmospheric pressure

\rho is the density

g is the acceleration due to gravity (Take g = 9.8 m/s^{2} )

h is the height

From the question

h = 30.0 m

\rho = 1.00 × 10³ kg/m³ = 1000 kg/m³

P_{o} = 101.3 kPa = 101300 Pa

Using the formula

P = P_{o} + \rho gh

P = 101300 + (1000×9.8×30.0)

P = 101300 + 294000

P =395300 Pa

P = 395.3 kPa

Hence, the absolute pressure at the bottom of the freshwater lake is 395.3 kPa

(b)

For the force exerted

From

P = F/A

Where P is the pressure

F is the force

and A is the area

Then, F = P × A

Here, The area will be area of the window of the underwater vehicle.

Diameter of the circular window = 34.1 cm = 0.341 m

From Area = πD²/4

Then, A = π×(0.341)²/4 = 0.0913269 m²

Now,

From F = P × A

F = 395300 × 0.0913269

F = 36101.5 N

Hence, the force exerted by the water on the window is 36101.5 N

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